# Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain

## Theorem

Let $\tuple {X, d}$ be a metric space.

Let $\tuple {Y, d'}$ be a complete metric space.

Let $A \subseteq X$.

Let $f : A \to Y$ be a uniformly continuous function.

Then there exists a unique continuous function $g : A^- \to Y$ such that:

- $\map g a = \map f a$

for all $a \in A$, where $A^-$ denotes the topological closure of $A$.

Furthermore, $g$ is uniformly continuous.

## Proof Outline

Suppose we had a uniformly continuous function $g$ satisfying the hypotheses.

We know that from Sequential Continuity is Equivalent to Continuity in Metric Space, that if $\sequence {a_n}$ is a sequence in $A$ converging to $a \in A^-$, we must have:

- $\map g {a_n} \to \map g a$

Since $\sequence {a_n}$ is a sequence in $A$, this is equivalent to:

- $\map f {a_n} \to \map g a$

So our first goal is to "fill in" the missing values by considering limits of sequences of the form $\sequence {\map f {a_n} }$ where $a_n \to a$.

To ensure uniqueness, we show that the limit is the same regardless of the choice of $\sequence {a_n}$.

We show that the thus constructed function is uniformly continuous, and so is our desired function.

## Proof

### Existence

Note that if $A$ is closed, then from Set is Closed iff Equals Topological Closure, we have:

- $A^- = A$

So taking $g = f$ suffices in this case.

Suppose now that $A$ is not closed.

#### Lemma 1

Let $\sequence {a_n}$ be a sequence in $A$ convergent to $a \in A^-$.

Then $\sequence {\map f {a_n} }$ converges.

$\Box$

We now show that the limit of $\sequence {\map f {a_n} }$ is independent of the particular $\sequence {a_n}$ chosen.

#### Lemma 2

Let $\sequence {a_n}$ be a convergent sequence in $A$.

Then the limit of $\sequence {\map f {a_n} }$ is dependent only on the limit of $\sequence {a_n}$.

That is, there exists a function $L : A^- \to Y$ such that:

- $\ds \lim_{n \mathop \to \infty} \map f {a_n} = \map L {\lim_{n \mathop \to \infty} a_n}$

for every convergent sequence $\sequence {a_n}$.

$\Box$

Now, define the function $g : A^- \to Y$ by:

- $\map g a = \begin{cases}\map L a & a \in A^- \setminus A \\ \map f a & \text{otherwise}\end{cases}$

We now show this function is uniformly continuous.

Let $\epsilon > 0$.

We want to show that there exists some $\delta' > 0$ such that whenever $a, b \in A^-$ and:

- $\map d {a, b} < \delta'$

we have:

- $\map {d'} {\map g a, \map g b} < \epsilon$

From Point in Closure of Subset of Metric Space iff Limit of Sequence:

- there exists a sequence $\sequence {a_n}$ in $A$ such that $a_n \to a$

and:

- there exists a sequence $\sequence {b_n}$ in $A$ such that $b_n \to b$.

We then have, by the Triangle Inequality:

- $\map {d'} {\map g a, \map g b} \le \map {d'} {\map g a, \map f {a_n} } + \map {d'} {\map f {a_n}, \map f {b_n} } + \map {d'} {\map f {b_n}, \map g b}$

Recall by construction of $g$:

- $\map f {a_n} \to \map g a$

and:

- $\map f {b_n} \to \map g b$

From the definition of convergence in a metric space, we can find $N_1 \in \N$ such that:

- $\map {d'} {\map g a, \map f {a_n} } < \epsilon/3$

for $n > N_1$.

We can also find $N_2 \in \N$ such that:

- $\map {d'} {\map g b, \map f {b_n} } < \epsilon/3$

for $n > N_2$.

We now wish to find $N_3 \in \N$ such that:

- $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/3$

for $n > N_3$.

Recall that since $f$ is uniformly continuous function, we can find $\delta > 0$ such that whenever:

- $\map d {a_n, b_n} < \delta$

we have:

- $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/3$

Then by the Triangle Inequality we have:

- $\map d {a_n, b_n} \le \map d {a_n, a} + \map d {a, b} + \map d {b, b_n}$

Since $a_n \to a$ we can pick $M_1 \in \N$ such that:

- $\map d {a_n, a} < \delta/3$

for $n > M_1$.

Since $b_n \to b$ we can pick $M_2 \in \N$ such that:

- $\map d{b, b_n} < \delta/3$

for $n > M_2$.

So, if:

- $\map d {a, b} < \delta/3$

and:

- $n > \max \set {M_1, M_2} = N_3$

we have:

- $\map d {a_n, b_n} < \delta$

and so:

- $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/3$

Now, let:

- $N = \max \set {N_1, N_2, N_3}$

Then for $n > N$ and whenever $\map d {a, b} < \delta/3$, we have:

- $\map {d'} {\map g a, \map g b} \le \map {d'} {\map g a, \map f {a_n} } + \map {d'} {\map f {a_n}, \map f {b_n} } + \map {d'} {\map f {b_n}, \map g b} < \epsilon$

So:

- $\map {d'} {\map g a, \map g b} < \epsilon$

whenever:

- $\map d {a, b} < \delta/3$

So $g$ is uniformly continuous.

From Uniformly Continuous Function is Continuous: Metric Space, $g$ is also continuous.

$\Box$

### Uniqueness

Let $h$ be another continuous function satisfying the hypotheses.

Then, for any $a \in A^-$ and sequence $\sequence {a_n}$ convergent sequence to $a$, we have:

- $\map h {a_n} \to \map h a$

and:

- $\map f {a_n} \to \map g a$

by Sequential Continuity is Equivalent to Continuity in Metric Space.

But since $h$ coincides with $f$ on $A$, this gives:

- $\map f {a_n} \to \map h a$

from Convergent Sequence in Metric Space has Unique Limit, we obtain:

- $\map h a = \map g a$

So we must have $g = h$.

$\blacksquare$